# In statistical mechanics, why do we consider number of states of a system in energy interval?

In statistical mechanics,when we go for calculating the no. of accessible micro states of a system, I notice that we always calculate the no. of micro states of that system in some energy interval say between $E$ and $E+dE$ instead of calculating the no. of micro states at particular energy $E$. Even,in micro canonical ensemble,in order to calculate the entropy of a system we use the renowned formula $$S=k \ln W$$ where $W$ is the number of accessible micro states of the system in energy interval between $E$ and $E+dE$ but not the micro states at energy $E$. So my question is why we always calculate micro states in enery range $dE$, but not at a particular energy $E$ of the system?

• Please don't shout at us. – Jon Custer Jan 24 '18 at 19:31
• If I randomly pick a number between 0 and 1, what is the probability that it will be, say, 0.5? – lemon Jan 24 '18 at 20:13

Disclaimer

I prefer the notation $E+\Delta E$, because I want it to be clear that we are talking about a finite quantity (i.e. nowhere we will take the limit $dE \to 0$ or such things).

Theoretical point of view

You can surely consider the states with energy precisely equal to $E$ and define the microcanonical probability density as

$$\rho(p,q) = C \cdot \delta(H-E)$$

where $C$ is a constant and $H=H(p,q)$ is the Hamiltonian. This is for example what Landau does in his book.

However, it can be shown that for large $N$ this definition is equivalent to

$$\rho(p,q)=C' \cdot \int_{E<H<E+\Delta E}d^{3N}pd^{3N}q$$

for a suitable constant $C'$ and for some constant $\Delta E$, which doesn't even necessarily be "small" (see for example my answer here, or Tuckerman's book).

Experimental point of view

There will always be some uncertainty on the measurements you can effectuate of the system's energy. If you want, this $\Delta E$ represents this uncertainty.

Therefore, while from a mathematical point of view it makes perfect sense to take $\delta(H-E)$ as your probability density, form an experimental point of view it makes much more sense to consider energies in some finite interval.

If all you want is to calculate the entropy, things get even "easier". Indeed, the high dimensionality ($6N$) of phase spaces makes it so that you can equivalently define the entropy as

$$S = k_B \log \int_{E<H<E+\Delta E}d^{3N}pd^{3N}q$$

$$S = k_B \log \int_{H<E}d^{3N}pd^{3N}q$$

or even

$$S = k_B \log \left( \partial_E \int_{H<E}d^{3N}pd^{3N}q\right)$$

where $\partial_E$ is the partial derivative with respect to $E$.

The difference between these definitions will be a constant of order $\log N$ or smaller, irrelevant since $S$ is of order $N$ (see Huang, Statistical Mechanics).

• @pragyanpratimbordoloi If this answer solved your doubt, you can consider accepting it. – valerio Feb 3 '18 at 10:35